288=64+48t+16t^2

Solution for 288=64+48t+16t^2 equation:

288=64+48t+16t^2

We move all terms to the left:

288-(64+48t+16t^2)=0

We get rid of parentheses

-16t^2-48t-64+288=0

We add all the numbers together, and all the variables

-16t^2-48t+224=0

a = -16; b = -48; c = +224;
Δ = b2-4ac
Δ = -482-4·(-16)·224
Δ = 16640
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{16640}=\sqrt{256*65}=\sqrt{256}*\sqrt{65}=16\sqrt{65}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{65}}{2*-16}=\frac{48-16\sqrt{65}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{65}}{2*-16}=\frac{48+16\sqrt{65}}{-32}
$